Help with impedance matching circuit design

I would imagine your video equipment is unbalanced (coax) whereas your ethernet cable is balanced, so you need a balun transformer with a 4:3 impedance ratio.

Are you sure your video equipment needs to be fed from a 75 ohm source? Ideally, it should be, but it's more important to terminate the transmission line.

You could possibly get away with a lossy-match by inserting a 25-ohm series resistor to terminate the source, but this would still leave the balanced / unbalanced problem. How about a 1:1 pulse transformer, followed by a 25-ohm series resistor on the output side? Being a lossy match, this would affect the amplitude of your video signal of course.

At the transmitting end, your video equipment will need to see a 75-ohm load. Again, this could be done with a lossy-match, but you'd need some amplification in there somewhere if you care about the amplitude of the video waveform. Another problem with a simple lossy match is that it only matches in one direction, so it won't absorb reflections going the other way.

Go to Google, and search on variations of broadband impedance matching balun transformer, and something is almost bound to float to the top. ;-)

Good Luck! Rich

Nope, actually this will be inside of a very small robot and I have already been told by the mechanical designers that I _cannot_ have any more wires due to space constraints. So I have to either use the existing wire assigned to ethernet (luckily I don't *always* need ethernet) or rig up some sort of wireless video feed to get it from point A to point B. I'm exploring both options at the moment so I need to prove/disprove that I can put a 75 Ohm video signal down a 100 Ohm cable and still have the picture be usable.

Thanks for the help guys!

-Will

Ok, after looking at a couple of suggestions here and other places I wound up trying a resistor matching circuit. Made up of three resistors on each side, joined together in the shape of pi (so extra points for geek factor) it worked like a champ for my test setup. Details below.

-Will

-----------+-----R2----+----------- | |

75 Ohm R1 R3 100 Ohm | | -----------+-----------+-----------

R1 = 61.5 Ohms R2 = 10 Ohms R3 = 82.1 Ohms

Hmmm- which design constraint is that? Video cable unavailable as a scarf-up item from the workplace?

Well, it depends on what you mean by "usable" picture. And the other issue is, if it's _inside_ the robot, it really shouldn't make that much difference how you wire it up, as long as you know to avoid ground loops and all that crap. 3' (1M) or so isn't long enough of a transmission to be so terribly worried about impedance matching, because at video frequencies the worst that could happen is a little loss of video resolution and probably virtually unnoticable ghosting.

But why do you need to transmit signals like that from one side of a robot to the other?

For that matter, why do you need ethernet from one side of a robot to the other?

I suspect there's something you're not telling us here.

Thanks, Rich

That circuit's not a match between 75 Ohm and 100 Ohm. Setting R1=150 Ohm and R2=50 Ohm and removing R3 will provide the minimum-loss resistive match between

75 Ohm and 100 Ohm.

--
Rick

That's not even close. You must have a funny idea of "matching." Remove R1 and R3, and make R2=25 ohms. Then the video amplitude will be exactly right(in most cases), and the return loss termination will be exactly

100 ohms (in most cases). The receive end should be terminated in 100 ohms total using external resistor and receiver input impedance.

How do you figure that? That will give -6dB at the R1/R2 junction and an additional 3.5dB to the line. The total amplitude will be -3.5dB down assuming the usual +6dB gain up internally for the 75 ohm series termination drive. The return termination is right.

Obviously wherever there exists a match there will be a point somewhere in the circuit at which the voltage is 6dB down on the unloaded generator voltage. If you wish to deliver the full power available from the generator into the load then the generator's voltage needs to be transformed losslessly, which is impossible with a resistive network, and requires either lumped elements or transmission line to achieve. That's why I said that this was the minimum loss topology for a RESISTIVE match.

Cheers, Rick